Question

ABC is an equilateral triangle with sides equal to 2 cm. BC is extended its own length to D, and E is the midpoint of AB. ED meets AC at F. Find the area of the quadrilateral BEFC in square centimeters in simplest radical form.

Answer 1

`(2\sqrt3)/(3)`
`cm^2`

Explanation:

Since
`\triangle ABC` is equilateral triangle, then
`\angle A = \pi /3`

So,

`S_(\triangle ABC) = (2 * 2 * sin(\pi /3))/(2) = (2 * 2 * √(3)/2)/(2) = √(3)`
`cm^2`

Then we need to find
`S_(\triangle AEF)` which can be computed by finding length_AF.

Let's call x = length_AF.

By Menelao's Theorem,

`(BE*x*CD)/(AE*(2-x)*BD) = (1*x*2)/(1*(2-x)*4)  = 1`

⇒ x = 4/3 cm

Thus,

`S_(\triangle AEF) = (1 * x * sin(\pi /3))/(2) = (1 * 4/3 * √(3)/2)/(2) = 1/√(3)`
`cm^2`

To find the area of quadrilateral
`BEFC`, we have to subtract
`S_(\triangle AEF)` from
`S_(\triangle ABC)`

Hence,

`S_(BEFC) = S_(\triangle ABC) - S_(\triangle AEF) = \sqrt3 -1/\sqrt3 = (2\sqrt3)/(3)`
`cm^2`